If the equation has two unequal real roots, then:
△=(k+ 1)^2-8(k+3)>; 0
k^2+2k+ 1-8k-24>; 0
k^2-6k-23>; 0
k^2-6k+9>; 32
(k-3)^2>; 32
K-3 > 4 √ 2 or k-3
K > 3+4 √ 2 or k < 3-√ 32
Let the two roots of the equation be a and b.
According to Vieta theorem
a+b=(k+ 1)/2
ab=(k+3)/2
According to the meaning, |a-b|= 1
So:
(a-b)^2= 1
(a+b)^2-4ab= 1
(k+ 1)^2/4-4*(k+3)/2= 1
(k+ 1)^2-8(k+3)=4
k^2+2k+ 1-8k-24-4=0
k^2-6k-27=0
(k+3)(k-9)=0
K=-3 or k=9
2. Substitute a=6-b into c2=ab-9 to obtain,
C2 = a b-9 =(6-b)b-9 = 6 b-B2-9 =-(b-3)2
∫C2≥0 and -(b-3)2≥0,
∴c=0 and b-3=0, which means c=0 and b=3.
∴a=6-b=6-3=3.